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Total Elastic Potential Energy 2

  1. Jan 27, 2005 #1
    I need to know how to go about finding the speed.

    A block B of mass 5 kg is fastened to one end of each of two springs. The other ends of the springs are attached to fixed points A and C, 4 metres apart on a smooth horizontal surface, as shown in the diagram.

    Spring AB has natural length 2 metres and modulus of elasticity 30 N, while BC has natural length 1 metre and modulus 40 N (you may assume that the springs meet at the centre of B).

    If the block is moved 0.5 metres towards C from its equilibrium position and then released, determine its speed as it passes through its equilibrium position.

    this is what i have tried

    Energy in string when moved 0.5m = total mech energy before the 0.5 movement
    E= 30*(0.5)^2/(2*2) = 1.575 J

    KE= 1/2*5*v^2 = 5/2*v^2 J and
    E in string 30*(3/8)^2/2*2 = 7.93 J

    7.93 + 5/2v^2 = 1.875
    this leads me to taking square root of a -ve number

    could someone help me!

    thanks in advance
  2. jcsd
  3. Jan 28, 2005 #2
    I'm going to assume this is a follow-up to the question I did before.

    The springs already had an extension at the equilibrium position, so when you move it box towards C, the extension will change.
    The extension in AB (x) is now: 0.5 + 8/11 = 27/22
    The extension in BC (y) is now: 3/11 - 0.5 = -5/22 (i.e. compression)

    Calculate the total mechanical energy at that point (which is only EPE). This should equal the total mechanical energy at the equilibrium position (which is KE & EPE).

    It's important that you calculate the total energy. You shouldn't get the square root of a negative number. :smile:
  4. Jan 28, 2005 #3
    cheers i can dig that
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